Les S be a scheme. Does there exist a faithfully flat morphism T to S with T a locally Noetherian scheme?
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7$\begingroup$ No, there is not necessarily such a $T$. For instance, let $S$ be $\text{Spec}(A)$ where $A=k[[x_1,x_2,...]]$. If there is such a faithfully flat $T$, then there is a point $t$ of $T$ which maps to the closed point of $S$. Let $B$ be the local ring of $T$ at $t$. Then there is a faithfully flat local homomorphism $A\to B$. In particular, the induced map $m_A/m_A^2 \to m_B/m_B^2$ is injective. Since $m_A/m_A^2$ is infinite dimensional, so is $m_B/m_B^2$, contradicting that $B$ is Noetherian. $\endgroup$– Jason StarrCommented Dec 15, 2011 at 16:27
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3$\begingroup$ @Jason: This is not a comment - it is an answer. $\endgroup$– Martin BrandenburgCommented Dec 15, 2011 at 19:52
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3$\begingroup$ @Jason: If $A \to B$ is a local extension of DVR's (automatically faithfully flat), then the map $m_A/m_A^2 \to m_B/m_B^2$ may be zero (e.g. if there is ramification). $\endgroup$– Akhil MathewCommented Dec 16, 2011 at 1:55
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$\begingroup$ @Akhil: You are right. That map may be zero. But the map from $m_A/m_A^2$ to $m_AB/m_A^2B$ is nonzero, which gives the same result. $\endgroup$– Jason StarrCommented Dec 17, 2011 at 19:02
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$\begingroup$ @Akhil: I should say a bit more. Since $B$ is $A$-flat, the induced map $B\otimes_A (m_A/m_A^2) \to m_AB/m_A^2B$ is an isomorphism. In particular, $m_AB/m_AB^2$ is a free $B/m_AB$-module of infinite rank. But that implies that $m_AB$ cannot be a finitely generated ideal in $B$, thus $B$ is not Noetherian. $\endgroup$– Jason StarrCommented Dec 17, 2011 at 19:41
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Note that if $T\to S$ is also quasicompact, then $S$ must be locally noetherian: this boils down to the well-known fact that if $A\to B$ is a faithfully flat ring homomorphism and $B$ is noetherian, then so is $A$.
This proves that Jason's example above is indeed a counterexample. More generally, any non-noetherian local scheme $S$ is a counterexample: if $T\to S$ is faithfully flat, there is an open affine $U\subset T$ which covers $S$.
EDIT: In fact, here is a complete answer ($S$ is any given scheme):
(1) The following are equivalent:
(1a) There exists a locally noetherian scheme $T$ and a faithfully flat and quasicompact morphism $T\to S$.
(1b) $S$ is locally noetherian.
(2) The following are equivalent:
(2a) There exists a locally noetherian scheme $T$ and a faithfully flat morphism $T\to S$.
(2b) For each $s\in S$, the ring $\mathcal{O}_{S,s}$ is noetherian.
Proof: exercise. To show that (2b) implies (2a), take $T=\coprod_{s\in S}\mathrm{Spec}(\mathcal{O}_{S,s})$.
answered Dec 16, 2011 at 8:06